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The correct OPTION is (b) sec^-1

The EXPLANATION: Space velocity has the dimensions of time^-1. A space velocity of 2 hr^-1 implies that 2 reactor volumes are fed PER hour.

Correct option is (a) –ri = \(\FRAC{1}{RT}\frac{dpi}{dt} \)

EXPLANATION: The design equation for batch SYSTEM is –ri = \(\frac{1}{V}\frac{dNi}{dt}.\) For constant VOLUME system

–ri = \(\frac{d(\frac{NI}{V})}{dt} = \frac{dCi}{dt}\) where C = p/RT. Hence, –ri = \(\frac{1}{RT}\frac{dpi}{dt}. \)

Right OPTION is (a) FA0\(\FRAC{DX}{dV}\) = -rA

For explanation: Molar flowrate and conversion are RELATED in the mole balance EQUATION to get the above equation.

The correct option is (B) \(\FRAC{Q}{V} \)

To elaborate: Space VELOCITY is the RATIO of volumetric flow rate to volume. It has the unit hr^-1.

Correct option is (C) Any arrangement

Easiest explanation: The FINAL concentration for both the CASES is same and can be calculated using the formula

CA2 = \(\frac{CA0}{(τ1K+1)(τ2k+1)} \)

Hence, it can be inferred that for first ORDER reactions, irrespective of the ARRANGEMENTS the final conversions are going to be the same.

Right choice is (a) \(\frac{VC_{A0}}{F_{A0}} \)

Easiest explanation: Space TIME is the RATIO of the product of reactor volume and initial concentration to the feed RATE. \(\frac{VC_{A0}}{F_{A0}} \)has the same unit as that of space time.

Right answer is (a) NA0\(\frac{dXA}{DT}\) = -rA V

Explanation: To obtain the TIME REQUIRED to achieve a CERTAIN conversion. Mole balance is WRITTEN in terms of conversion and then differentiated to get the above equation.

The correct option is (a) True

For explanation: Comparison of performance of SINGLE MFR and PFR are MADE for the NTH order and was inferred that the MFR is ALWAYS larger than a PFR for a given duty.

The CORRECT ANSWER is (d) 2

Explanation: RESIDENCE time = \(\frac{volume}{volumetric \, FLOW \, rate} = \frac{10}{5} \) = 2.

Correct option is (b) True

The EXPLANATION: For a FIRST order reaction, –\(\frac{dC_A}{DT} \) = kCA0

CA = CA0(1 – XA)

DIFFERENTIATING on either sides, dCA = -CA0dXA

-CA0 \(\frac{dXA}{dt} \) = kCA0(1 – XA)

\(\frac{dXA}{(1 – XA)} \) = kdt

-ln(1-XA) = kt

Hence, CONVERSION is not a function of CA0.

Right answer is (a) True

To EXPLAIN I would say: The composition at exit is identical to the composition in the REACTOR. Hence, REACTION rate is MEASURED at the exit CONDITIONS.

The CORRECT option is (c) same as

For explanation: For PFR, \(\frac{τ}{CA0} = \int_0^{XA}\frac{dXA}{-rA} \)

For Batch reactor, \(\frac{t}{CA0} = \int_0^{XA}\frac{dXA}{-rA} \)

 From the above equations it is CLEAR that THEORETICALLY the element of fluid reacts for same LENGTH of time in PFR and Batch reactor. But shutdown time as to be considered for actual DESIGN.

Right option is (c) Batch reactor

Easy explanation: When SMALL scale production is desired, HIGH or STRINGENT QUALITY STANDARDS to be met batch reactors used extensively.

Right option is (d) Jones

The best EXPLANATION: A graphical procedure was presented by Jones in 1951 to estimate the outlet composition of MFR’s in SERIES.

Correct choice is (c) t = NA0\(\int_0^{XA} \frac{dXA}{-r_{AV}} \)

Easiest explanation: The differential form of the DESIGN EQUATION is widely used in interpreting the lab data. Mole balance WRITTEN in TERMS of conversion is INTEGRATED to get the above equation.

Correct OPTION is (C) Any arrangement

Explanation: The final concentration for both the cases is same and can be calculated USING the formula

CA2 = \(\frac{CA0 e^{-kτp}}{(τmk+1)} \)

Therefore, any arrangement does not MAKE any difference for first order reactions.

Right option is (a) Time REQUIRED to process unit VOLUME of feed

To elaborate: SPACE time is the time spent by a unit volume of reactants inside the reactor. It is the performance MEASURE for a flow reactor.

The correct option is (b) 0.33

Easiest EXPLANATION: For a CSTR, the performance EQUATION is FA0XA = (-RA)V

t = \(\FRAC{C_{A0}- CA}{(-rA)} \)

rA = \(\frac{10-5}{100} \) = 0.05.

The correct answer is (b) 20

Explanation: FA0 XA = (-rA)V

(-rA) = kCA0(1 – XA)

500 × 0.5 = K × 50 × (1-0.5) × 0.5

k = 20.

The correct ANSWER is (d) 27

Easiest explanation: \(\frac{τ}{CA0} = \frac{XA}{-rA} = \frac{XA}{k(CA0(1-XA))^{1.5}} \)

τ = \(\frac{0.7*10}{0.15(10(1-0.7))^1.5} \)

τ = \(\frac{V}{V0} \) = 8.981 min

V = 8.98*3 = 26.94 LIT .

Right option is (c) W = FA0 \(\int_0^X \FRAC{dX}{-rA} \)

EXPLANATION: The general mole balance for flow REACTORS is modified to get the above equation. The reaction rate is EXPRESSED in mol/s.g of catalyst in this equation.

Right OPTION is (B) Plug, Small MFR, Large MFR

The best I can EXPLAIN: For this kind of rate-concentration curve, the concentration should be kept as high as possible, THEREFORE this ARRANGEMENT should be used.

Right answer is (b) 60.5

The EXPLANATION: For a second order reaction OCCURRING in a CSTR,

XA1 = (1-XA1)^2 V1

\(\frac{0.45}{0.90} = \frac{(1-0.45)^2 V1}{(1-0.9)^2 V2} \)

\(\frac{V_1}{V_2} \) = 0.0165.

The correct choice is (a) For pure gaseous REACTANTS, the best reactor SUITED for carrying out reaction is CSTR.

Easy explanation: For pure gaseous reactants, \(\frac{C_p}{-∆H_r} \) is small, where Cp is the specific HEAT capacity of the gas and ∆Hr is the ENTHALPY change of reaction. CSTR gives a high conversion for small values of \(\frac{C_p}{-∆H_r}. \)

The correct option is (b) V = FA0 \(\int_0^X\frac{dX}{-RA} \)

EXPLANATION: To DETERMINE the volume of PFR required to obtain a certain conversion, the MOLE balance equation in terms of molar flowrate and conversion is written and INTEGRATED to get the above equation.

Right choice is (a) Space VELOCITY is the NUMBER of unit volumes of FEED that can be treated PER unit time.

The explanation: Space velocity is the INVERSE of space time. It gives a measure of what volume of feed can be processed in a unit time.

The correct ANSWER is (b) 0.75

The EXPLANATION is: -LN(1-XA) = kt

XA = 0.75.

Right choice is (a) \(\frac{τ}{CA0} = \frac{XA}{-rA} \)

Easy explanation: General design EQUATION of CSTR is MODIFIED and WRITTEN in terms of INITIAL concentration and space time to get the above equation.

The correct option is (c) Takes 10 mins to treat ONE reactor volume of feed

To ELABORATE: TIME required to process one reactor volume of feed is called SPACE time.

The correct option is (c) 2

To explain: -RA = k(Csub>A)^n is the GENERAL RATE expression

Ln(-rA ) = lnk+ nlnCA

The slope of the line gives the order of the reaction.

The correct option is (a) -RA = 0.17 CA

For explanation I would say: 2A → R

εA = \(\frac{.7-1}{1}\) = -0.3

\(\begin{array}{c|c c}

& X_A=0 & X_A = 1 \\

\hline

 A & 0.6 & 0.3 \\

 I & 0.4 & 0.4 \\

\hline

 & 1 & 0.7 \\

\END{array}

\)

ΔV = V-Vo

= Vo – 0.15 VO – VO = -0.15 VO

-ln(1-\(\frac{ΔV}{Vo εA}\)) = kt

=> k = 0.173 min^-1.

The correct ANSWER is (a) \(\FRAC{mol^2}{m^6} \) and min

To explain: Two QUANTITIES of the same DIMENSIONS can only be added. As the unit of concentration is \(\frac{mol^2}{m^3} \) , k1 has the units \(\frac{mol^2}{m^6} \).

Equating the units of all known quantities on both sides, \(\frac{mol}{m^3 min} = \frac{mol^2}{m^6} × \frac{1}{k_2×(\frac{mol}{m^3})}. \)

Hence, k2 has the unit of min.

Correct option is (a) True

For explanation I would say: Density change for most liquid PHASE REACTIONS are negligible. Hence, constant VOLUME BATCH reactors are widely USED.